Does this $\max$ mean that we need to maximize the regret in this regret formula?
Yes, you're interpreting the $\max$ there wrongly. In your second formula
$$
\operatorname{Regret}_{T}(\mathcal{H})=\max _{h^{\star} \in \mathcal{H}} \operatorname{Regret}_{T}\left(h^{\star}\right) \label{1}\tag{1}
$$
The sign $=$ means &quot;is defined as&quot;, so maybe the following notation is less confusing
$$
\operatorname{Regret}_{T}(\mathcal{H}) \triangleq \max _{h^{\star} \in \mathcal{H}} \operatorname{Regret}_{T}\left(h^{\star}\right)
$$
In fact, in section 2.1 of the same paper, there's a similar but more detailed formula that should clarify the meaning of $\max$
$$
\operatorname{Regret}_{T}(\mathcal{H})=\max _{h \in \mathcal{H}}\left(\sum_{t=1}^{T}\left|p_{t}-y_{t}\right|-\sum_{t=1}^{T}\left|h\left(\mathbf{x}_{t}\right)-y_{t}\right|\right)
$$
Note that $\operatorname{Regret}_{T}(\mathcal{H})$ is defined for $\mathcal{H}$, while your first formula is defined for $x$. So, \ref{1} is the regret for the whole hypotheses class $\mathcal{H}$, which is thus the maximum regret that you can have across all possible hypotheses $h \in \mathcal{H}$. This should make sense.